Phasor-diagram Solution

Figure 7.11 has two panels that together show how the individual phasor voltages in a series LCR circuit add up to the source voltage.

Panel (a) shows the four phasors — current $I$, resistor voltage $V_R$, inductor voltage $V_L$, and capacitor voltage $V_C$ — drawn on a set of $x$–$y$ axes. The key phase relationships are:

• $I$ and $V_R$ are parallel (pointing up and to the right), because the voltage across a resistor is in phase with the current.
• $V_L$ points up and to the left, leading $I$ by $\pi/2$ (90°).
• $V_C$ points straight down, lagging $I$ by $\pi/2$ (90°), exactly opposite to $V_L$.
• A small right-angle square at the origin marks the axes.

This panel establishes the relative directions of the phasors before any addition is done.

Panel (b) shows the vector addition that leads to the source voltage $V$. The phasor $V_R$ is drawn along the current direction. The two reactive phasors $V_L$ and $V_C$ are combined into a single phasor $(V_C + V_L)$; because they point in opposite directions, the resultant is a shorter phasor pointing down and to the right (the direction of the larger of the two). The resultant of $V_R$ and $(V_C + V_L)$ is the source phasor $V$, which forms the hypotenuse of a right triangle. The phase angle $\phi$ is marked between $V_R$ and $V$. The dashed top segment is labelled $v_{Cm} - v_{Lm}$, the horizontal side is $v_{Rm}$, and the hypotenuse is $v_m$.

Physical idea: In a series LCR circuit, the same current flows through all three elements, but the voltages across them are not in phase. The resistor voltage is in phase with the current, the inductor voltage leads by 90°, and the capacitor voltage lags by 90°. Because $V_L$ and $V_C$ are exactly opposite, they partially cancel. The net voltage across the reactive part is $|V_C - V_L|$, and the total source voltage is the vector sum of $V_R$ and this net reactive voltage.

Key formulas developed from this figure:

The Pythagorean theorem applied to the right triangle in panel (b) gives the amplitude of the source voltage:

Substituting $v_{Rm} = i_m R$, $v_{Cm} = i_m X_C$, and $v_{Lm} = i_m X_L$ (from Eq. 7.22) yields:

This leads to the definition of impedance $Z$:

so that $i_m = v_m / Z$ (Eq. 7.25b). …

The impedance diagram (Fig. 7.12) is a right-angled triangle that summarises the relationship between resistance, reactance, and impedance in a series LCR circuit. It is not a phasor diagram (which shows time-varying voltages and currents), but a static geometric representation of the magnitudes of the circuit's opposition to alternating current.

• The horizontal base is labelled $R$ — the resistance (in ohms). This side is drawn along the real axis.
• The vertical side (right side of the triangle) is labelled $X_C - X_L$ — the net reactance. This is the difference between capacitive reactance $X_C = \frac{1}{\omega C}$ and inductive reactance $X_L = \omega L$. The vertical direction represents the imaginary axis.
• The hypotenuse is labelled $Z$ — the impedance (in ohms). It runs from the bottom-left origin to the top-right vertex.
• A small right-angle square is drawn at the bottom-right vertex, confirming the triangle is right-angled.
• The angle $\phi$ is marked at the bottom-left vertex, between the base $R$ and the hypotenuse $Z$. This angle is the phase difference between the source voltage and the circuit current.

Physical idea: The diagram teaches that impedance $Z$ is the vector sum of resistance $R$ (along the real axis) and net reactance $(X_C - X_L)$ (along the imaginary axis). The Pythagorean theorem gives the magnitude of $Z$, and the angle $\phi$ gives the phase relationship.

Key formulas developed from this figure:

where:

• $Z$ = impedance of the series LCR circuit (Ω)
• $R$ = resistance (Ω)
• $X_C = \frac{1}{\omega C}$ = capacitive reactance (Ω)
• $X_L = \omega L$ = inductive reactance (Ω)

$\tan \phi = \frac{X_C - X_L}{R}$ …

What Figure 7.13 Shows

The figure has two panels, both for the case $X_C > X_L$ — meaning the capacitive reactance dominates over inductive reactance, so the circuit behaves as predominantly capacitive.

Panel (a): Phasor diagram

• A phasor I (current) is drawn at some angle.
• A phasor V (source voltage) is drawn at a smaller angle — specifically, V points up-and-right at angle $\omega t_1$.
• Since $X_C > X_L$, the current leads the voltage: the I phasor is drawn ahead (higher angle) than V.
• The phase angle $\phi$ is marked between V and I, showing how much the current leads the voltage.
• The phasors rotate anticlockwise with angular frequency $\omega$; at the instant $t_1$, their positions are frozen.

Panel (b): Graphs of $v$ and $i$ versus $\omega t$

• Horizontal axis: $\omega t$ (angular time), with ticks at $0$, $\omega t_1$, $\pi$, $2\pi$.
• Vertical axis: instantaneous voltage $v$ (solid curve) and current $i$ (dashed curve).
• The current curve leads the voltage curve: the dashed peak occurs before the solid peak by a horizontal offset equal to the phase angle $\phi$.
• This offset is marked near the peaks, showing that $i$ reaches its maximum earlier than $v$.

Physical Idea

The figure visually demonstrates that in a series LCR circuit, when $X_C > X_L$, the net reactance is capacitive, so the current leads the source voltage by a phase angle $\phi$. The phasor diagram shows the steady-state relationship between the phasors, while the time graphs show how this phase difference appears in the actual sinusoidal waveforms.

Key Formulas Developed with This Figure

From the textbook analysis using the phasor diagram (Fig. 7.11 and 7.12), the following results are obtained:

Impedance

where $R$ is resistance, $X_C = \frac{1}{\omega C}$ is capacitive reactance, and $X_L = \omega L$ is inductive reactance. …